Answer:
[tex]4.5\times 10^{14} N[/tex]
Explanation:
[tex]q_1=60 C[/tex]
[tex]q_2=-12 C[/tex]
[tex]q_3=7 C[/tex]
Distance between q1 and q2,d=20 cm
[tex]r=\frac{d}{2}=\frac{20}{2}=10 cm=10\times 10^{-2} m[/tex]
1m =100 cm
Electric force on charge q3 due to charge q1
[tex]F_1=\frac{kq_1q_3}{r^2}=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}[/tex] (away from q1)
Electric force on charge q3 due to charge q2
[tex]F_2=\frac{kq_2q_3}{r^2}=\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}[/tex](attract toward q2)
Net force=[tex]F=F_1+F_2[/tex]
[tex]F=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}+\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}[/tex]
[tex]F=\frac{9\times 10^9\times 7}{(10\times 10^{-2})^2}(60+12)[/tex]
[tex]F=4.5\times 10^{14} N[/tex]
